Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her next N (1 ≤ N ≤ 1,000,000) hours (conveniently labeled 0..N-1) so that she produces as much milk as possible. Farmer John ha…

Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her next N (1 ≤ N ≤ 1,000,000) hours (conveniently labeled 0..N-1) so that she produces as much milk as possible. Farmer John ha…

#include <iostream> #include <cstdio> #include <algorithm> using namespace std; struct cow { int start; int endd; int price; }; bool cmp(cow a,cow b) { return a.start<b.start; } int main() { int n,m,r; cow a[1005]; cin>>n>>…

Milking Time 直接翻译了 Descriptions 贝茜是一个勤劳的牛.事实上,她如此​​专注于最大化她的生产力,于是她决定安排下一个N(1≤N≤1,000,000)小时(方便地标记为0..N-1),以便她生产尽可能多的牛奶. 农民约翰有一个M(1≤M≤1,000)可能重叠的间隔列表,他可以在那里进行挤奶.每个区间我有一个起始小时(0≤starting_houri≤N),一个结束小时(starting_houri <ending_houri≤N),以及相应的效率(1≤efficien…

区间按左端点排序,dp. #include<cstdio> #include<algorithm> using namespace std; #define N 1001 struct Point{int l,r,w;}a[N]; bool operator < (const Point &a,const Point &b){return a.l<b.l;} int n,m,K,f[N]; int main() { scanf("%d%d%d&…

#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <cstdlib> using namespace std; int N, M, R; // N个小时, M个时间间隔, R个休息时间 + ; struct Cow { int start; int end; int value; Cow(, , ) : start(s),…

                                                                                             Milking Time Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8048   Accepted: 3388 Description Bessie is such a hard-working cow. In fact, she i…

Description Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her next N (1 ≤ N ≤ 1,000,000) hours (conveniently labeled 0..N-1) so that she produces as much milk as possible. Fa…

POJ3176 Cow Bowling 题意 输入一个n层的三角形,第i层有i个数,求从第1层到第n层的所有路线中,权值之和最大的路线. 规定:第i层的某个数只能连线走到第i+1层中与它位置相邻的两个数中的一个. 思路 最显而易见的是使用二维数组动态规划计算. 比如dp[i][j]表示以第i行j列的位置作为终点的路线的最大权值. (注意区分初始化时的意义) 那么dp[i][j]的最大值取决于dp[i-1][j-1]和dp[i-1][j],从这两者之间筛选出最大值,加到dp[i][j]上,即为dp…

传送门: http://poj.org/problem?id=3616 Milking Time Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13406   Accepted: 5655 Description Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she dec…